Prove that $a_{n+1} = a_n^2 - a_n+1$ for all $n$ large enough 
Prove that if $a_1,a_2,\ldots$ are positive integers so that $a_{n+1}\ge \prod_{k=1}^n a_k$ and $\sum_{n=1}^\infty \frac{1}{a_n}$ is rational, $a_{n+1} = a_n^2 - a_n+1$ for all sufficiently large $n$.

I don't have a concrete idea of what to do; maybe a proof by contradiction could work? Would solving the following simpler problem be of any use: given $a_1 = 2, a_{n+1} = a_n^2 - a_n+1, n>0$, prove that if $m\neq n, a_m$ and $a_n$ are coprime and $\sum_{i=1}^\infty \frac{1}{a_i} = 1$.
The difficulty with this problem is that I'm not sure how I can make use of the given conditions.
 A: This exact problem is a theorem proved by Erdos and Straus back in 1963 in a paper called: On the irrationality of certain Ahmes series. I'm having some trouble finding a copy of the original paper online, so here is a paper by Badea which starts with their theorem and then expands on it.

EDIT: Thank you @Will_Jagy for finding a copy of the original Erdos/Straus paper.
A: This is only a partial solution, of the "simpler problem" (and it's simple, indeed): Let $a_1=2$ and $a_{n+1}=a_n^2-a_n+1$ for $n\ge1$. Then,
$$a_{n+1}-1=a_n\,(a_n-1)\tag{1}.$$ From this, we can show by induction that
$$a_{n+1}-1=(a_1-1)\prod^n_{m=1}a_m\tag{2}.$$ So $a_m$ is a divisor of $a_{n+1}-1$ for $m\le n$, i.e. $a_{n+1}$ and $a_m$ are coprime. Moreover, (1) can be witten as
$$\frac1{a_n}=\frac1{a_n-1}-\frac1{a_{n+1}-1},$$ so we have the telescoping series
$$\sum^\infty_{n=1}\frac1{a_n}=\frac1{a_1-1}=1.$$
The original question does not seem to be entirely hopeless, considering some expansions studied by Waclaw Sierpinski.
EDIT: there is some sort of expansion related (but, alas, not identical) to the problem at hand. Let's assume we have some real, positive value $x=x_0$, and let's define $a_k$ as the least positive integer so that $1/a_k<x_k$, and $x_{k+1}=x_k-1/a_k$. Since we always have $x_n>0$ (a not so greedy algorithm), this is always an infinite process, but still, $$x=\sum^\infty_{k=0}\frac1{a_k}.$$ Due to the minimality of $a_k$, we have
$$\frac1{a_k}<x_k\le\frac1{a_k-1},$$ and thus
$$0<x_k-\frac1{a_k}=x_{k+1}\le\frac1{a_k\,(a_k-1)}.$$ This implies $$a_{k+1}>a_k\,(a_k-1)\tag{3}.$$ It's easy to see that (3) makes the expansion
$$x=\sum^\infty_{k=0}\frac1{a_k}$$ unique. And, lo and behold, $x$ is rational if and only if $a_{k+1}=a^2_k-a_k+1$ for $k\ge K$: if that's so, according to the above, $$\sum^\infty_{k=K}\frac1{a_k}=\frac1{a_K-1}$$ is rational, and so is
$$\sum^{K-1}_{k=0}\frac1{a_k}+\frac1{a_K-1}.$$ On the other hand, if $x_k=m/n$ is rational and a reduced fraction, it's clear that $n=m\,q-r$ with $0<r<m$, meaning $$x_{k+1}=\frac{m}{n}-\frac1{a_k}=\frac{m}{n}-\frac1q=\frac{r}{n\,q}.$$ Even if this is a reduced fraction as well (in reality, cancelations occur), $r<m$, so we'll inevitably arrive at some $x_k=1/d$. And then, $a_k=d+1$, $$x_{k+1}=\frac1{d\,(d+1)}=\frac1{a_k\,(a_k-1)}$$ and thus $a_{k+1}=a_k\,(a_k-1)+1$, and so on.
